Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}2x-3y &= -3 \\ 5x-8y &= -8\end{align*}$
Begin by moving the $y$ -term in the second equation to the right side of the equation. $5x = 8y-8$ Divide both sides by $5$ to isolate $x$ $x = {\dfrac{8}{5}y - \dfrac{8}{5}}$ Substitute this expression for $x$ in the first equation. $2({\dfrac{8}{5}y - \dfrac{8}{5}}) - 3y = -3$ $\dfrac{16}{5}y - \dfrac{16}{5} - 3y = -3$ Simplify by combining terms, then solve for $y$ $\dfrac{1}{5}y - \dfrac{16}{5} = -3$ $\dfrac{1}{5}y = \dfrac{1}{5}$ $y = 1$ Substitute $1$ for $y$ in the top equation. $2x-3( 1) = -3$ $2x-3 = -3$ $2x = 0$ $x = 0$ The solution is $\enspace x = 0, \enspace y = 1$.